Question

Suppose that a balanced die is rolled repeatedly until the same number appears on two successive rolls, and let X denote the number of rolls that are required. Determine the value of Pr(X=x) for x=2,3,...

Answer 1

Well, what is Pr(X=2)? And Pr(X=3)?

Answer 2

what is a balanced die
that is a regular die right?

Answer 3

Its just a die that isn't loaded...and doesnt favor one number over the other...

Answer 4

Pr(X=2) = Pr(second roll = first roll) = 1/6

Now Pr(X=3) = Pr(2nd roll not = first roll).Pr(3rd roll=2nd roll)
= 5/6 . 1/6

Hence Pr(X=4) = ...

Answer 5

I like this question a lot because you'll see this:

Once you find the answer, you'll also be able to show that
\[\sum_{n} \Pr(X=n) \ = 1\]
which makes sense as the probability that infinity many rolls have no repeats is zero.

Answer 6

My friend got the answer:

(5/6)^(x-2) times (1/6)

Answer 7

I came to that solution I believe...but it's been taking me a long time to realize how I got it...

Answer 8

who cares. What's the logic for the answer? Follow what I wrote above and hopefully the first two cases make sense. What is Pr(X=4)?

Answer 9

(5/6)^2 (1/6) ?

Answer 10

Pr(X=4) = Pr(2nd roll not = 1st roll) . Pr(3rd roll not = 2nd roll) . Pr(4th roll = 3rd roll)
= 5/6 . 5/6 . 1/6

Answer 11

right. And now generalize

Answer 12

Oh thank you so very much! I understand the problem quite well now...or I hope I do. Are you a math major by the way?

Answer 13

was, yes

Answer 14

That was a very helpful way of showing me how to understand the problem, thank you again.